# $\lim_{a\to\infty} {a\int_{0}^{\frac{\pi}{4}}e^x\space \tan^a{x}\space dx}$

$$Let \space I(a) = \int_{0}^{\frac{\pi}{4}} {e^{x} \tan^{a} x} \space . \space Find \space \lim_{a\to\infty}aI(a).$$

Well this is an integral that I'm curently dealing with and so far I've tried to solve it using different approaches. Still, I haven't solved it and what I find always is an undefined answer. Even when I used mathematica it returned an undefined answer. This solution in my opinion is most likely to be the one that works: $$\\$$ Our Integral is $$\int_{0}^{\frac{\pi}{4}}e^x\space \lim_{a\to\infty}[{a\tan^a{x}}]\space dx,$$ letting $$A=\lim_{a\to\infty}{a\tan^a{x}}$$ We know that $$x \in [0,\frac{\pi}{4}]\space \Rightarrow \space \tan{x} \in[0,1] \space \Rightarrow \space \lim_{a\to\infty}a\tan{^ax}= \ \begin{cases} 0 & 0\leq x< \frac{\pi}{4} \\ 1 & x=\frac{\pi}{4}\\ \end{cases} \$$ Which tells us that $$A = \ \begin{cases} 0 \times \infty & 0\leq x< \frac{\pi}{4} \\ \infty & x=\frac{\pi}{4}\\ \end{cases}$$ Which are some undefined integrands... . Mathematica says the integral is undefined but the book, Advanced Calculus Explored, says it has an answer. I think this is the right solution and I just need to work on the limit a little bit more, but this is something that I'm stuck in. I appreciate any kind of hints or helps.

• $\lim_{0 \to \infty}$? It is not always valid to say $\lim \int = \int \lim$. Jun 24 at 12:30
• @Saman Look up the dominated convergence theorem.
– Gary
Jun 24 at 12:50
• Sorry I don't know what it is Jun 24 at 12:56
• Integration by parts gives $I(a) = \exp(\pi/4) - a[I(a-1)+I(a+1)]$, so if the limit exists, it's $\frac{1}{2} \exp(\pi/4)$. Now maybe to find some good squeeze functions? Jun 24 at 13:03
• @Saman That is why I told you to look it up.
– Gary
Jun 24 at 13:24

Our problem is a disguised form of the result

$$\tag 1 \lim_{a\to \infty} a\int_0^1 y^a f(y)\,dy=f(1),$$

which holds for any continuous $$f$$ on $$[0,1].$$ It has been proved on MSE many times. The proof is quite simple for $$f$$ continuously differentiable on $$[0,1],$$ using integration by parts.

To see why the above will help us in the question at hand, let $$x=\arctan y.$$ The original expression then turns into

$$a\int_0^1 y^a e^{\arctan y}\frac{1}{1+y^2}\,dy.$$

Letting $$f(y) = e^{\arctan y}\frac{1}{1+y^2},$$ we see that $$(1)$$ implies our limit is

$$f(1)= e^{\arctan 1}\frac{1}{1+1^2} = \frac{e^{\pi/4}}{2}.$$

• +1 for the simpler approach. Jun 24 at 23:26
• This solution is awesome! I tried to prove it using only integration by parts but I couldn't; by this I mean only with the knowledge of calculus, not real analysis or ... because I haven't studied it yet. Is there any proof without using rigorous analysis? Because I didn't find any of this kind in even the MSE Jun 25 at 13:36
• @Saman The proof of $(1)$ is fairly simple if $f$ is continuously differentiable. Did you try that? Once we have that, there's nothing going on except for the substitution $x=\arctan y.$
– zhw.
Jun 27 at 14:47
• Thanks. Yes I got it Jun 28 at 11:27

Rewrite the integral as

$$\int_0^{\pi/4} dx \, \exp{x} \, \exp{\left [a \left (\log{\tan{x}} \right ) \right ]}$$

Sub $$x=\pi/4-y$$ and use the tangent addition rule and the integral is

$$\exp{\left ( \frac{\pi}{4} \right )} \int_0^{\pi/4} dy \, \exp{(-y)} \, \exp{\left [a \left (\log{\left (\frac{1-\tan{y}}{1+\tan{y}} \right )} \right ) \right ]}$$

Not that the dominant contribution to the integral as $$a \to \infty$$ is in the interval in a small neighborhood about $$y=0$$. So expand the integrand about $$y=0$$; then the interval of integration may be expanded out to $$[0,\infty)$$ because the other contributions are exponentially subdominant. Accordingly, as $$a \to \infty$$, the integral behaves as

$$\exp{\left ( \frac{\pi}{4} \right )} \int_0^{\infty} dy \, e^{-2 a y} = \frac1{2 a} \exp{\left ( \frac{\pi}{4} \right )}$$

Accordingly, the sought-after limit is $$\frac12 \exp{\left ( \frac{\pi}{4} \right )}$$

This has been verified in Mathematica.

• Thank you sooooo much! It's exactly the answer of the book Jun 24 at 13:12
• @Saman: pro-tip: when trying it out in Mathematica, evaluate the limit numerically; that is, compute a table of values for larger and larger $a$. Jun 24 at 13:13
• It might clarify things a bit to make explicit that for $y$ near 0, you're expanding $\log \left( \frac{1-\tan y}{1+\tan y} \right) \sim -2y$. Jun 24 at 18:56
• This is a special case of Laplace's method (the general method, not the particular set of cases called Laplace's method in the Wikipedia article). Jun 24 at 18:57
• @DanielSchepler: I intended to outline just a few steps and give a sense of the solution. I leave minutiae like that to the reader. Jun 24 at 19:00

Another approach (that works for this type of limits, with $$a$$ times an integral of an $$a$$-th power times something else) is to "absorb" the factor $$a$$ using integration by parts. Let $$g_a(x)=\tan^a x$$. Then $$a\cdot e^x\tan^a x=f(x)g_a'(x)$$, where $$f(x)=e^x\tan x\cos^2 x\color{LightGray}{[{}=\ldots]}$$. Hence $$aI(a)=\int_0^{\pi/4}f(x)g_a'(x)\,dx=\underbrace{f(\pi/4)g_a(\pi/4)}_{=(1/2)e^{\pi/4}}-\underbrace{f(0)g_a(0)}_{=0}-\int_0^{\pi/4}f'(x)g_a(x)\,dx.$$ The last integral tends to $$0$$ as $$a\to\infty$$ (by DCT as noted in comments if you know it, or by bounding the integrand from above: $$f'(x)$$ by a constant, and $$g_a(x)$$ by, say, $$\tan x\leqslant 4x/\pi$$).